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Find an equation for the tangent to the curve at $P\left( \dfrac{\pi}{2},3 \right )$ and the horizontal tangent to the curve at $Q.$ $$y=5+\cot x-2\csc x$$

$y\prime=-\csc ^2 x -2(-\csc x \cot x)$
$y\prime= 2\csc x \cot x - \csc ^2 x\implies$ This is the equation of the slope.

Now I find the slope at $x=\dfrac{\pi}{2}:$
$$y^{\prime}=2\csc \left( \frac{\pi}{2} \right) \cot \left( \frac{\pi}{2} \right)- \csc ^2 \left( \frac{\pi}{2} \right)$$ $$y^{\prime}= -1$$

Equation for the tangent to the curve at $P$:
$$y-3=-1\left( x-\frac{\pi}{2} \right)$$ $$y=-x+\frac{6+\pi}{2}$$

Then I found the horizontal tangent at $Q$, which is where the slope is $0.$ So I set the slope equation equal to $0:$

$$y^{\prime}= 2\csc x \cot x - \csc ^2 x=0$$ $$-\csc ^2 x = -2\csc x \cot x$$

$$-\frac{1}{\sin^2 x}=-2\left( \frac{1}{\sin x} \frac{\cos x}{\sin x} \right )$$

$$\cos x = \frac{1}{2}$$


I don't know how to proceed from here. Since $\cos x = \dfrac{1}{2}$, $\cos^{-1}\left (\dfrac{1}{2}\right)=\dfrac{\pi}{3}$.
Plugging this into the equation would give the $y$ value, and I'd be able to find the equation of the tangent line, but I'm confused because there is not just one $x$ to consider, since $x=2n \pm \dfrac{\pi}{3}$.

The answer for equation of tangent at $Q$ is $y=5-\sqrt{3}$, and I don't know how to get this.

Can you please show how to work this last part in details? Thank you.

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  • $\begingroup$ $x=2n\pi\pm\pi/3$ $\endgroup$ – Mikasa Dec 27 '13 at 17:43
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You need to find the $y$ coordinate of $Q$, let it be called $y_0$. The horizontal tangent passes from $Q=(x_0,y_0)$ and so the equation will be $y=y_0$.

$\sin(x_0)=\sqrt{1-\cos^2(x_0)}=\dfrac{\sqrt{3}}2$ since $\cos(x_0)=\dfrac{1}2$, so $\cot(x_0)=\dfrac{1}{\sqrt{3}}$.
And $\csc(x_0)=\dfrac{2}{\sqrt{3}}$ so $y_0=5-\sqrt{3}$

Even if you choose some other $x_0$, you will get the same $y_0$. It's a horizontal tangent line.

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I think what's you need is to consider $$\sin(2k \pi-\alpha)=-\sin(\alpha),~~~\sin(2k \pi+\alpha)=\sin(\alpha)\\ \cot(2k \pi-\alpha)=-\cot(\alpha),~~~\cot(2k \pi+\alpha)=\cot(\alpha)$$

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  • $\begingroup$ @B.S. Happy new year my friend $\endgroup$ – Adi Dani Dec 30 '13 at 9:16
  • $\begingroup$ @AdiDani: However my year will start within Spring, thanks for saying so my dear friend. :-) $\endgroup$ – Mikasa Dec 30 '13 at 9:19
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The first part looks good. For the second part as you see the solution to the equation gives $$2 \cot x \csc x - \csc^2 x = 0 \Rightarrow x=2n \pi \pm\frac{\pi}{3}.$$

Choose $x=\frac{\pi}{3}$ as it is one such solution. We find the point on $y$ to be $$5+\cot\left( \frac{\pi}{3} \right)-2\csc\left( \frac{\pi}{3} \right)=5-\sqrt{3},$$ And so $y=5-\sqrt{3}$ is indeed a horizontal tangent line, in particular the horizontal tangent line at $x=\frac{\pi}{3}$. You can verify this in WA.

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Here is the graph. Do you see what happens?

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  • $\begingroup$ thanks, yes, I know that the horizontal tangent will be a constant, however I don't know how to show that it will be $y=5-\sqrt{3}$ $\endgroup$ – Emi Matro Dec 27 '13 at 17:42
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This means that there can be infinitely many horizontal tangents. As simple as that .

Just pick up the one you want.

$x = 2n \pm \dfrac{\pi}3$ is one solution.

Corresponding $y$ value

$y = 5 + \cot\left(2n \pm \dfrac{\pi}3\right) -2\csc\left(2n \pm \dfrac{\pi}3\right)$

$y = 5 \pm \dfrac1{\sqrt3} \pm \dfrac4{\sqrt3} $

$y = 5 \pm \sqrt3$

So one solution is $ y = 5 - \sqrt3$

It's better you solve $x = 2n + \dfrac{\pi}3$ and $x = 2n - \dfrac{\pi}3$ separately so as to not get yourself confused.

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